P.E. of a certain spring when stretched from natural length through a distance 0.3, m is 10, J . amo

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5.Work, Energy, Power and Collision

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The $P.E.$ of a certain spring when stretched from natural length through a distance $0.3\, m$ is $10\, J$. The amount of work in joule that must be done on this spring to stretch it through an additional distance $0.15\, m$ will be ................ $\mathrm{J}$

A

$10$

B

$20$

C

$7.5$

D

$12.5$

Solution

The initial potential energy of the spring is given as,

$U_{i}=\frac{1}{2} k x^{2}$

$10=\frac{1}{2} k(0.3)^{2}$

$k=\frac{20}{0.09}$

The final potential energy of the spring is given as,

$U_{f}=\frac{1}{2} k x_{1}^{2}$

$=\frac{1}{2} \times \frac{20}{0.09}(0.45)^{2}$

$=22.5 \mathrm{J}$

The amount of work done is given as,

$W=U_{f}-U_{i}$

$=22.5-10$

$=12.5 \mathrm{J}$

Thus, the amount of work done is $12.5 \mathrm{J}$.

Standard 11

Physics

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